The sequence of Fibonacci numbers is defined by the
initial values f0 = 0, f1 = 1, and the recurrence fn
= fn-1 + fn-2. The first several values are

Notice that the even-ordered terms satisfy the relations

and so on. Similarly the odd-ordered terms satisfy the
relations

and so on. One way of proving that this pattern continues
is by induction. We know by inspection up to a certain n that consecutive
Fibonacci numbers are given by

(Note that this requires f1 = f2 =
1.) Adding these together, term by term, it follows from the basic Fibonacci
recurrence that the next even-ordered value in the sequence is

which is of the same form as the previous even-ordered
term, but with n+1 in place of n. Likewise the next odd-ordered value in the
sequence can be found by adding this expression for f2(n+1) to the
preceding expressing for f2n+1, which gives

where we have used the fact that f1 = 1. This
is of the same form as the previous odd-ordered term, but with n+1 in place
of n, so the induction is complete. There, the interleaving odd and even
parts of the Fibonacci numbers satisfy the relations

Naturally these identities are closely related to the
generating functions for the recurrence relations. We know the ratios of
successive values of fn asymptotically approach successive powers of the
dominant root f of the characteristic
equation x2 – x – 1 = 0. Therefore, with n = 6, the even-ordered
relation approaches

and if we divide this by f10
we get

Taking this to the limit as n goes to infinity, and
substituting for the derivative of the geometric series, we get

which is equivalent to the characteristic equation f2 – f – 1 = 0. However, this isn’t the only possible expansion of
the terms of the even (or odd) ordered Fibonacci sequences. For example,
given the even-ordered sequence 0, 1, 3, 8, 21, 55, 144, … we could simply
apply the “greedy algorithm” by expressing each term as the sum of the
maximum number of the immediately preceding terms. Thus we have

From this it’s easy to see that f2n = f2(n-1)
– f2(n-2) + 2f2(n-1), so the even-ordered Fibonacci
numbers satisfy the recurrence

which has the characteristic equation

This shows that another way of arriving at the recurrence
for the kth Fibonacci numbers is to solve for the (k-2)th degree polynomial
that gives, when multiplied by the characteristic polynomial of the Fibonacci
sequence, a polynomial whose only non-zero coefficients are for powers that
are multiples of k. For example, to find the recurrence for every third
Fibonacci number, we evaluate the product

We need the coefficients of x5, x4,
x2, and x to vanish, so we require A = 1, B = 2, C = -1, and D = 1. Hence the characteristic
polynomial for every third Fibonacci number is

with the corresponding recurrence relation

Naturally the same type of relations can be derived for
any linear recurring sequence. For example, consider the “tribonacci”
sequence, defined by the initial values t0 = t1 = t2
= 1, and the recurrence tn = tn-1 + tn-2 + tn-3.
The first several values are shown below.

To find the recurrence formula for every third term of
this sequence, we can apply the greedy algorithm to express the terms 1, 9,
57, 355, 2209, 13745, 85525, … as follows

From this it’s easy to see that t3n = t3(n-1) – 5t3(n-2) + t3(n-3) + 6t3(n-1), so we have the recurrence

corresponding to the characteristic polynomial

We can relate this back to the greedy recurrence by
examining the asymptotic expression in terms of the dominant root f, which can be written in the form

This implies that f is a root of the characteristic
polynomial, as expected. Incidentally, the “poles” of this recurrence are 0
and the cube roots of 1.